Q. When 9.65 C of electricity is passed through a solution of silver nitrate (atomic weight of Ag = 107.87 taking as 108), the amount of silver deposited is
A) 5.8 mg
B) 10.8 mg
C) 15.8 mg
D) 20.8 mg
Explanation
(b)
$w_{Ag} = \frac{E_{Ag} \times \ Q}{96500} = \frac{108 \times 9.65}{96500} = 1.08 \times 10^{- 2}\ g = 10.8\ mg$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.